You have found the following ages (in years) of all 5 bears at your local zoo: $ 64,\enspace 46,\enspace 2,\enspace 64,\enspace 11$ What is the average age of the bears at your zoo? What is the variance? You may round your answers to the nearest tenth.
Solution: Because we have data for all 5 bears at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $5$ ages and divide by $5$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{64 + 46 + 2 + 64 + 11}{{5}} = {37.4\text{ years old}} $ Find the squared deviations from the mean for each bear. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $64$ years $26.6$ years $707.56$ years $^2$ $46$ years $8.6$ years $73.96$ years $^2$ $2$ years $-35.4$ years $1253.16$ years $^2$ $64$ years $26.6$ years $707.56$ years $^2$ $11$ years $-26.4$ years $696.96$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{707.56} + {73.96} + {1253.16} + {707.56} + {696.96}} {{5}} $ $ {\sigma^2} = \dfrac{{3439.2}}{{5}} = {687.84\text{ years}^2} $ The average bear at the zoo is 37.4 years old. The population variance is 687.84 years $^2$.